## Thinking Mathematically (6th Edition)

We can find the total patient load. total load = 119 + 165 + 216 = 500 We can find the standard divisor. $standard~divisor = \frac{total~load}{number~of~ doctors}$ $standard~divisor = \frac{500}{10}$ $standard~divisor = 50$ The standard divisor is 50. We can find the standard quota for each clinic. Clinic A: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{119}{50}$ $standard~quota = 2.38$ Clinic B: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{165}{50}$ $standard~quota = 3.30$ Clinic C: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{216}{50}$ $standard~quota = 4.32$ If each clinic is apportioned its lower quota, the number of doctors apportioned is 2 + 3 + 4, which is 9 doctors. Since there are a total of 10 doctors available, there is one surplus doctor. To obtain a sum of 10 doctors, we need to find a modified divisor that is slightly less than the standard divisor. Let's choose a modified divisor of 43. Note that it may require a bit of trial-and-error to find a modified divisor that works. We can find the modified quota for each clinic. Clinic A: $modified~quota = \frac{patient~load}{modified~divisor}$ $modified~quota = \frac{119}{43}$ $modified~quota = 2.77$ Clinic B: $modified~quota = \frac{patient~load}{modified~divisor}$ $modified~quota = \frac{165}{43}$ $modified~quota = 3.84$ Clinic C: $modified~quota = \frac{patient~load}{modified~divisor}$ $modified~quota = \frac{216}{43}$ $modified~quota = 5.02$ Using Jefferson's method, the modified quota is rounded down to the nearest whole number. Each clinic is apportioned the following number of doctors: Clinic A is apportioned 2 doctors. Clinic B is apportioned 3 doctors. Clinic C is apportioned 5 doctors. Note that the total number of doctors apportioned is 10, so using a modified divisor of 43 is acceptable.